Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 7 / 4 / i / Solution

Put $w(x)=(\sin x/x{)}^2$, with $w(0)=1$, and $M={\sup }_r|A_r|<\infty$. Since $0\le w\le 1$, the finite-prefix bound is $\left|{\sum }_{r=1}^n{A}_{r}w(rk)\right|\le nM$. For $k\ne 0$, $w(rk)\le 1/(r^2{k}^2)$ gives absolute convergence of the tail and the correct estimateThe PDF omits the factor $k^{-2}$ from this bound. Its printed version is false: choose $A_4=1$ and all other $A_r=0$, take $n=3$, and let $k\to 0$. The left side tends to one while the printed right side is $2/3$. Thus the intended bound must be qualified as above. At $k=0$, even absolute convergence need not hold for a general sequence tending to zero.

For each fixed $n$, the displayed finite-prefix limit follows immediately: $|k{\sum }_{r=1}^n{A}_{r}w(rk)|\le |k|nM\to 0$. The stronger statement needed for part (ii) isHere is a proof, so the fixed-prefix formula is not silently substituted for an infinite-series result. Set $N=\lfloor 1/|k|\rfloor$ for small nonzero $k$. The multiplied tail is at most $2{\sup }_{r>N}|A_r|/(N|k|)$, which tends to zero because $N|k|\to 1$. For the prefix, choose $J$ with $|A_r|<\eta$ for $r>J$. ThenIts upper limit is at most $\eta$, and $\eta$ is arbitrary. This proves sinc-squared summation of a sequence tending to zero, including approach from either sign of $k$.

For the unheaded piecewise-affine continuation, continuity gives $Ac+B=A^{\prime }c+B^{\prime }$. Evaluate the asserted symmetric quotient at $t=c$. For sufficiently small positive $h$,Its zero limit forces $\boxed{A=A^{\prime }}$, and the continuity relation then forces $\boxed{B=B^{\prime }}$. Only the condition at the corner is needed; at points away from it such a quotient is automatically zero even for unequal slopes.